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Q: What does A-Level Physics: 13) Thermodynamic Systems Guide cover? A: Internal energy, pΔV work, the Zeroth & First Laws, plus specific heat and latent heat - this guide unpacks Topic 13 of the 2026 H2 Physics syllabus.
TL;DR Thermodynamics is the energy balance sheet of the universe. Nail the distinctions between U, Q, W and you will unlock marks in gas laws, kinetic theory and every practical calorimetry task.
Stay synced with the adjacent thermal-physics chapters via the H2 Physics notes hub; it bundles Temperature & Ideal Gases plus exam drills referenced here.
1 Internal energy (U)
The macroscopic state of a system fixes its internal energy: the sum of random microscopic kinetic (Ek) and potential (Ep)
energies. No single thermometer can reveal
U
directly - you must track
how
energy enters or leaves.
1.1 Parent takeaway
A student who writes “heat stored in the gas” is forfeiting method marks. Heat is a transfer, not a store.
1.2 Mini-drill
State whether each contributes to U:
Scenario
Contributes to U?
Translational motion of the whole cylinder
No
Vibrations between gas molecules
Yes
2 Thermodynamic temperature (T)
Absolute temperature in kelvin is directly proportional to the mean kinetic energy per particle. Doubling T doubles ⟨Ek⟩. Always quote exam answers to 2-3 s.f. unless otherwise stated.
3 Heating (Q) vs work (W)
Symbol
Process
Positive when...
Typical formula
Q
Energy transfer by heating
Energy enters system
Q=mcΔT
W
Mechanical work
Done on system
W=−pΔV (constant p; ΔV=Vf−Vi
Sign hack: In SEAB mark schemes, work done by the gas is negative.
4 Zeroth law - the thermometer principle
If A is in thermal equilibrium with B, and B with C, then A is in equilibrium with C. This justifies using a third body (a thermometer) to compare temperatures.
5 First law - the energy ledger
ΔU=Q+W.
Positive Q: heating adds energy.
Positive W: surroundings compress system.
For an ideal gas in free expansion, Q=0 and W=0, hence ΔU=0: temperature unchanged.
6 Specific heat capacity (c)
Defined as the heat required per unit mass per kelvin rise:
Q=mcΔT.
6.1 Quick-check
Water has c≈4.18kJ⋅kg−1⋅K−1 - roughly 11x that of copper (≈0.39kJ⋅kg−1⋅K−1). That is why water is a coolant.
7 Specific latent heat (L)
Energy per unit mass for a phase change at constant T:
Q=mL.
Common values:\
Substance
Lf (fusion) / kJ⋅kg−1
Lv (vaporisation) / kJ⋅kg−1
Water
334
2260
8 WA & Paper 4 timing rules
Use syllabus pacing as a guide: Paper 2/3 average ~1.6 min/mark; Paper 4 ~3 min/mark.
Copy units first when using data-book tables.
Show full working; SEAB awards method marks even if arithmetic slips.
Need structured practice on Thermodynamic Systems? Our H2 Physics tuition programme covers this topic with weekly problem sets and Paper 4 practical drills.
Comprehensive revision pack
9478 Section III, Topic 13 Syllabus outcomes at a glance
Outcome (a) - explain internal energy and thermodynamic temperature.
Outcome (b) - apply the first law of thermodynamics to closed systems.
Outcome (c) - calculate heating and work done in simple processes.
Outcome (d) - determine energy changes using specific heat capacity and latent heat.
Outcome (e) - interpret p-V diagrams and identify isothermal, adiabatic, isobaric, and isochoric processes.
Concept map (in words)
Identify the process path (constant T, V, p or adiabatic). Track energy transfers: heating (Q) and work (W) alter internal energy (U). Use c and L for temperature changes and phase changes respectively. p-V diagrams visualise work as area under curve.
Key relations
Quantity/Process
Expression / reminder
First law
ΔU=Q+W (W positive when done on system)
Work at constant pressure
W=pΔV (positive if compression)
Specific heat
Q=mcΔT
Specific latent heat
Q=mL
Isothermal ideal gas
pV=constant, ΔU=0
Isochoric process
ΔV=0⇒W=0
Adiabatic (ideal gas)
Q=0⇒ΔU=W
Internal energy of ideal gas
U=23nRT (monatomic)
Derivations & reasoning to master
First law sign convention: derive for compression/expansion, noting SEAB's “work done by gas is negative”.
Work on a p-V diagram: show W=∫pdV and interpret the area under the curve.
Heating vs temperature change: explain why adding latent heat occurs without temperature change.
Adiabatic vs isothermal: compare outcomes for ΔU to emphasise physical differences.
Worked example 1 - heating and work
An ideal gas is compressed slowly from 4.0L to 2.0L at constant pressure 150kPa while 1.6kJ of heat is removed. Find the change in internal energy.
Method: W=−pΔV (constant p); apply ΔU=Q+W. Discuss the sign of ΔU (gas cooled).
ΔV=2.0−4.0=−2.0L⇒W=−pΔV=−150(−2.0)=+300J=0.30kJ.
Heat is removed so Q=−1.6kJ. Hence
ΔU=Q+W=(−1.6+0.30)kJ=−1.3kJ.
Worked example 2 - latent heat application
0.45kg of ice at −12∘C is heated to steam at 110∘C. Calculate the total energy required. Use cice, cwater, csteam, Lf, Lv.
Approach: break into stages: warm ice → melt → warm water → vaporise → superheat steam. Sum each Q.
Using typical data-book values cice≈2.1kJ⋅kg−1⋅K−1, cwater≈4.18kJ⋅kg−1⋅K−1, csteam≈2.0kJ⋅kg−1⋅K−1, Lf=334kJ⋅kg−1, Lv=2260kJ⋅kg−1:
Parents: schedule a 1-hour Thermodynamic Systems clinic two weeks before WA 2.
Students: print the First Law equation and stick it on your water bottle - then apply it to every past-year Qn you meet.
Last updated 14 Jul 2025. Next review: upon release of the 2027 draft syllabus.
